Calculate joint Distribution function (X,Y), when X~Unif(0,1), Y=X

Question

I am new to this topic right now. This is how I proceeded-

$\int_{0}^{x}\int_{0}^{x} 1 dy dx$ = ${x^2}\over{2}$

Also, we were supposed to calculate its mixed partials?

For that $\partial^2{{x^2}\over{2}}\over\partial{x}\partial{y}$ = 1 for x between [0,1].

I don't know if I am proceeding right or not. Is my solution correct?

Answer 1

If $z_1$ or $z_2$ are negative, then $Pr(X \le z_1, Y \le z_2)=0$.

Let $z_1, z_2 $ are positive, then

\begin{align} Pr(X \le z_1, Y \le z_2) &= Pr(X \le \min(z_1, z_2))\\ &= \min(z_1, z_2,1) \end{align}